A mixture of 0.2063 mol of Cl2, 0.2093 mol of H2O, 0.1887 mol of HCl, and...

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A mixture of 0.2063 mol of Cl2, 0.2093 mol of H2O, 0.1887 mol ofHCl, and 0.09871 mol of O2 is placed in a 1.0-L steel pressurevessel at 575 K. The following equilibrium is established: 2 Cl2(g)+ 2 H2O(g) 4 HCl(g) + 1 O2(g) At equilibrium 0.05821 mol of O2 isfound in the reaction mixture. (a) Calculate the equilibriumpartial pressures of Cl2, H2O, HCl, and O2. (b) Calculate KP forthis reaction.

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4.1 Ratings (691 Votes)

Â Â Â Â Â Â Â Â Â Â Â Â Â Â 2 Cl2(g) + 2 H2O(g) ----> 4 HCl(g) + 1 O2(g)

initialÂ Â Â Â Â Â Â 0.2063 molÂ Â 0.2093 molÂ Â Â 0.1887 mol 0.09871 mol

changeÂ Â Â Â Â Â Â 2*0.0405 mol 2*0.0405 mol 4*0.0405 mol 0.0405 mol
Â Â Â Â Â
equilibriumÂ Â Â 0.2873 molÂ Â Â 0.2873 molÂ Â 0.0267 mol 0.05821 mol

at equilibrium

ntotal = 0.2873 +0.2873 +0.0267 +0.05821 = 0.66 mol

Ptotal = ntotal*RT/V

Â Â Â Â Â Â = 0.66*0.0821*575/1

Â Â Â Â Â Â = 31.16 atm

partial pressure of Cl2(pCl2) = (ncl2/ntotal)*ptotal

Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â = 0.2873/0.66*31.16
Â Â Â Â Â Â Â Â
Â Â Â Â Â Â Â Â Â Â Â = 13.56 atm
Â Â Â pH2O = 13.56 atm

Â Â pHCl = (0.0267/0.66)*31.16 = 1.26 atm

Â Â pO2 = (0.05821/0.66)*31.16 = 2.75 atm

Â Â Kp = pHCl^4*pO2 / pCl2^2*pH2O^2

Â Â Â Â Â Â = (1.26^4*2.75/(13.56^2*13.56^2))

Kp = 2.05*10^-4

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